In Exercises 9–16, find a basis for the eigenspace corresponding to each listed eigenvalue.

15.\(A = \left( {\begin{array}{*{20}{c}}4&2&3\\{ - 1}&1&{ - 3}\\2&4&9\end{array}} \right)\),\(\lambda = 3\)

Eigenvalue: Let \(\lambda \) is a scaler, \(A\) is an \(n \times n\) matrix and \({\bf{x}}\) is an eigenvector corresponding to \(\lambda \), \(\lambda \) is said to an eigenvalue of the matrix \(A\) if there exists a nontrivial solution \({\bf{x}}\) of \(A{\bf{x}} = \lambda {\bf{x}}\).

Eigenvector: For a \(n \times n\) matrix \(A\), whose eigenvalue is \(\lambda \), the set of a subspace of \({\mathbb{R}^n}\) is known as an eigenspace, where a set of the subspace of is the set of all the solutions of \(\left( {A - \lambda I} \right){\bf{x}} = 0\).

The given matrix is \(A = \left( {\begin{array}{*{20}{c}}4&2&3\\{ - 1}&1&{ - 3}\\2&4&9\end{array}} \right)\), where \(\lambda = 3\).

As, \(\lambda = 3\) are the eigenvalue of the matrix \(A\), so they satisfy the equation \(A{\bf{x}} = \lambda {\bf{x}}\).

For \(\lambda = 3\), solve \(\left( {A - \lambda I} \right){\bf{x}} = 0\), for which first evaluate \(\left( {A - \lambda I} \right)\).

\(\begin{array}{c}\left( {A - 3I} \right) = \left( {\begin{array}{*{20}{c}}4&2&3\\{ - 1}&1&{ - 3}\\2&4&9\end{array}} \right) - 3\left( {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}4&2&3\\{ - 1}&1&{ - 3}\\2&4&9\end{array}} \right) - \left( {\begin{array}{*{20}{c}}3&0&0\\0&3&0\\0&0&3\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}1&2&3\\{ - 1}&{ - 2}&{ - 3}\\2&4&6\end{array}} \right)\end{array}\)

Write the obtained matrix in the form of an augmented matrix, where for \(A{\bf{x}} = 0\), the augmented matrix given by \(\left( {\begin{array}{*{20}{c}}A&0\end{array}} \right)\).

\(\left( {\begin{array}{*{20}{c}}1&2&3&0\\{ - 1}&{ - 2}&{ - 3}&0\\2&4&6&0\end{array}} \right)\)

The obtained matrix is not in a reduced form, so reduce it in row echelon form by applying row operations.

\(\begin{gathered} \hfill \left( {\begin{array}{*{20}{c}} 1&2&3&0 \\ { - 1}&{ - 2}&{ - 3}&0 \\ 2&4&6&0 \end{array}} \right)\xrightarrow{{{R_2} \to {R_2} + {R_1}}}\left( {\begin{array}{*{20}{c}} 1&2&3&0 \\ 0&0&0&0 \\ 2&4&6&0 \end{array}} \right) \\ \hfill \xrightarrow({}){{{R_3} \to {R_3} - 2{R_1}}}\left( {\begin{array}{*{20}{c}} 1&2&3&0 \\ 0&0&0&0 \\ 0&0&0&0 \end{array}} \right) \\ \end{gathered} \)

Write a system of equations corresponding to the obtained matrix.

\(\begin{array}{c}{x_1} + 2{x_2} + 3{x_3} = 0\\{x_2},{x_3},{\rm{ free variables}}\end{array}\)

As \({x_2},{x_3}\) are free variables, then;

\({x_1} = - 2{x_2} - 3{x_3}\)

So, the general solution is given as:

\(\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right) = {x_2}\left( {\begin{array}{*{20}{c}}{ - 2}\\1\\0\end{array}} \right) + {x_3}\left( {\begin{array}{*{20}{c}}{ - 3}\\0\\1\end{array}} \right)\)

So, \(\left\{ {\left( {\begin{array}{*{20}{c}}{ - 2}\\1\\0\end{array}} \right),\left( {\begin{array}{*{20}{c}}{ - 3}\\0\\1\end{array}} \right)} \right\}\) is the basis for the eigenspace for \(\lambda = 3\).